# Print the degree of every node from the given Prufer sequence

• Difficulty Level : Hard
• Last Updated : 22 Jun, 2022

Given a Prufer sequence, the task is to find the degrees of all the nodes of the tree made by the prufer sequence.
Examples:

```Input: arr[] = {4, 1, 3, 4}
Output: 2 1 2 3 1 1

The tree is:
2----4----3----1----5
|
6

Input: arr[] = {1, 2, 2}
Output: 2 3 1 1 1```

A simple approach is to create the tree using the Prufer sequence and then find the degree of all the nodes.
Efficient approach: Create a degree[] array of size 2 more than the length of the prufer sequence, since the length of prufer sequence is N – 2 if N is the number of nodes. Initially, fill the degree array with 1. Iterate in the Prufer sequence and increase the frequency in the degree table for every element. This method works because the frequency of a node in the Prufer sequence is one less than the degree in the tree.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to print the degrees of every``// node in the tree made by``// the given Prufer sequence``void` `printDegree(``int` `prufer[], ``int` `n)``{``    ``int` `node = n + 2;` `    ``// Hash-table to mark the``    ``// degree of every node``    ``int` `degree[n + 2 + 1];` `    ``// Initially let all the degrees be 1``    ``for` `(``int` `i = 1; i <= node; i++)``        ``degree[i] = 1;` `    ``// Increase the count of the degree``    ``for` `(``int` `i = 0; i < n; i++)``        ``degree[prufer[i]]++;` `    ``// Print the degree of every node``    ``for` `(``int` `i = 1; i <= node; i++) {``        ``cout << degree[i] << ``" "``;``    ``}``}` `// Driver code``int` `main()``{``    ``int` `a[] = { 4, 1, 3, 4 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);``    ``printDegree(a, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `    ``// Function to print the degrees of every``    ``// node in the tree made by``    ``// the given Prufer sequence``    ``static` `void` `printDegree(``int` `prufer[], ``int` `n)``    ``{``        ``int` `node = n + ``2``;` `        ``// Hash-table to mark the``        ``// degree of every node``        ``int``[] degree = ``new` `int``[n + ``2` `+ ``1``];` `        ``// Initially let all the degrees be 1``        ``for` `(``int` `i = ``1``; i <= node; i++)``        ``{``            ``degree[i] = ``1``;``        ``}` `        ``// Increase the count of the degree``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``degree[prufer[i]]++;``        ``}` `        ``// Print the degree of every node``        ``for` `(``int` `i = ``1``; i <= node; i++)``        ``{``            ``System.out.print(degree[i] + ``" "``);``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `a[] = {``4``, ``1``, ``3``, ``4``};``        ``int` `n = a.length;``        ``printDegree(a, n);``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python3 implementation of the approach` `# Function to print the degrees of``# every node in the tree made by``# the given Prufer sequence``def` `printDegree(prufer, n):`` ` `    ``node ``=` `n ``+` `2` `    ``# Hash-table to mark the``    ``# degree of every node``    ``degree ``=` `[``1``] ``*` `(n ``+` `2` `+` `1``)` `    ``# Increase the count of the degree``    ``for` `i ``in` `range``(``0``, n):``        ``degree[prufer[i]] ``+``=` `1` `    ``# Print the degree of every node``    ``for` `i ``in` `range``(``1``, node``+``1``): ``        ``print``(degree[i], end ``=` `" "``)``     ` `# Driver code``if` `__name__ ``=``=` `"__main__"``:`` ` `    ``a ``=` `[``4``, ``1``, ``3``, ``4``]``    ``n ``=` `len``(a)``    ``printDegree(a, n)` `# This code is contributed by Rituraj Jain`

## C#

 `// C# implementation of the approach``using` `System;``    ` `class` `GFG``{` `// Function to print the degrees of every``// node in the tree made by``// the given Prufer sequence``static` `void` `printDegree(``int` `[]prufer, ``int` `n)``{``    ``int` `node = n + 2;` `    ``// Hash-table to mark the``    ``// degree of every node``    ``int``[] degree = ``new` `int``[n + 2 + 1];` `    ``// Initially let all the degrees be 1``    ``for` `(``int` `i = 1; i <= node; i++)``    ``{``        ``degree[i] = 1;``    ``}` `    ``// Increase the count of the degree``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``degree[prufer[i]]++;``    ``}` `    ``// Print the degree of every node``    ``for` `(``int` `i = 1; i <= node; i++)``    ``{``        ``Console.Write(degree[i] + ``" "``);``    ``}``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]a = {4, 1, 3, 4};``    ``int` `n = a.Length;``    ``printDegree(a, n);``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`2 1 2 3 1 1`

Time Complexity: O(N), as we are using a loop to traverse N times. Where N is the number of elements in the array.

Auxiliary Space: O(N), as we are using extra space degree array.

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